Class Session 04: Errors and fluctuations

Errors and Central limit theorem

Consider a direct sampling simulation, the outcome is usually a sequence of values
$(\xi_1, \xi_2, \ldots, \xi_N)$
which are different realizations of a random variable ξ , of distribution π(ξ). We assume that these values are independent and identically distributed (i.i.d.).

Using these N realisations, the best estimation of the mean
$\langle \xi\rangle$
of the distribution π(ξ) of ξ is
$X_N=\frac{1}{N}\sum_{i=1}^N \xi_i .$
Can we determine the statistical error associated with this estimation?

The mean XN being the average of N i.i.d. random variables, the central limit theorem should apply and the result writes:
$X_N=\langle \xi\rangle \pm \frac{\sqrt{\text{Var}(\xi_i)}}{\sqrt{N}}.$

Test this prediction with the following example: Sample N random numbers
$\{\xi_1,\xi_2,\ldots, \xi_N\}$
taken from a uniform distribution on [0,1] and give an estimation of the average with its associated error.

Quality Control

We consider the direct sampling of the distribution
$\pi(\xi) =\alpha/x^{\alpha+1} \quad\text{for}\quad x\in [1,+\infty[$

• How can you sample this distribution?
• Compute the average of ξ analytically.
• Estimate the error using the previous method. Compare your results for different values of α. (First take α=3 and then try α=1.2).

import math, numpy, random
Ntime = 500000
random.seed(2)
alpha = 3.
xi = []
for itime in range(Ntime):
    xi.append('think a little bit')
xerror = numpy.std(numpy.array(xi))/math.sqrt(Ntime)
xm = numpy.mean(numpy.array(xi))
print 'alpha = ',alpha
print 'exact mean', alpha / (alpha - 1.)
print 'simulation', xm,'+/-', xerror

To understand the difference between the different values of α let us introduce the "Quality control" method, which consists in plotting the running average defined as
$\frac{1}{i}\sum_{j=1}^i \xi_j\,.$

import pylab, math, numpy, random
Ntime = 10000
random.seed(2)
alpha = 3
xmeanexact = alpha / (alpha - 1.)
xicontrol = []
xtot = 0.
for itime in range(Ntime):
    xi = 'think a little bit'
    xtot += xi
    xicontrol.append( xtot / float(itime + 1))
pylab.title('Quality Control, alpha = 3')
pylab.xlabel('time')
pylab.ylabel('running average of xi')
time = [t for t in range(Ntime)]
meanexact = [ xmeanexact for t in range(Ntime)]
pylab.plot(xicontrol, 'r-')
pylab.plot(time, meanexact, 'k-')
pylab.axis([0, Ntime, xmeanexact * 0.9, xmeanexact * 1.1])
pylab.show()

• Run this program for Ntime=10000, 50000, 250000 and compare your results for different values of α.
• Compute analytically the variance of ξ as a function of α. Comment.

Stable distribution

Take Ntime=10,100,1000... Study the distribution of the variable
$\frac{1}{N}\sum_{i=1}^N \xi_i$
as N becomes large. Show that after a proper rescaling of this variable, the distribution converge (for large N) to a unique curve. What is this limit distribution?

 import pylab, math, numpy, random
 Nsample = 20000
 Ntime = 100
 random.seed(2)
 alpha = 3
 xmean = alpha/(alpha-1.)
 Nfactor = float(Ntime)**0.5   #(1. - 1./alpha)
 xsum = [0. for n in range(Nsample)]
 for isample in range(Nsample):
     xi = [0. for n in range(Ntime)]
     for itime in range(Ntime):
         xi[itime] = random.random()**(-1./alpha)
     xsum[isample]=(sum(xi)/float(Ntime) - xmean) * Nfactor
 pylab.hist(xsum, bins=30,range=(-10,10),normed=True)
 pylab.title('Example 3, Rescaled Distributions, alpha = 3')
 pylab.xlabel('distribution')
 pylab.ylabel('x')
 pylab.show()

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